Merge pull request #56 from stoyanK7/9.4_C-pointers_fix-typos-print-statements

fix typos in print statements

Author: syreal17

chapters/c.adoc

@@ -404,14 +404,14 @@ int main() {
  printf("\n This is arr[0]: %c ", arr[0]);
  printf("\n This is *arr: %c ", *(arr+0));
  //as well as:
- printf("\n This is arr[0]: %c ", arr[1]);
- printf("\n This is *(arr+0): %c ", *(arr+1));
- printf("\n This is arr[1]: %c ", arr[2]);
- printf("\n This is *(arr+1): %c ", *(arr+2));
- printf("\n This is arr[2]: %c ", arr[3]);
- printf("\n This is *(arr+2): %c ", *(arr+3));
- printf("\n This is arr[3]: %c ", arr[4]);
- printf("\n This is *(arr+3): %c ", *(arr+4));
+ printf("\n This is arr[1]: %c ", arr[1]);
+ printf("\n This is *(arr+1): %c ", *(arr+1));
+ printf("\n This is arr[2]: %c ", arr[2]);
+ printf("\n This is *(arr+2): %c ", *(arr+2));
+ printf("\n This is arr[3]: %c ", arr[3]);
+ printf("\n This is *(arr+3): %c ", *(arr+3));
+ printf("\n This is arr[4]: %c ", arr[4]);
+ printf("\n This is *(arr+4): %c ", *(arr+4));
  //understanding that, you can see now why in C, a thing that looks very weird as the following, makes sense:
  printf("\n This is 1[arr]: %c ", 1[arr]);
  //As you see, it printed 'e', because that expression is just *(1+a), which is the same as *(a+1)